Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
\;A){\rm{ }}{\left( {2x - 3} \right)^2} + \left( {3x + 1} \right)\left( {4 - 2x} \right)\\
= 4{x^2} - 12x + 9 + 12x - 6{x^2} + 4 - 2x\\
= - 2{x^2} - 2x + 13\\
{\rm{ }}B){\rm{ }}\left( {x + y} \right){\rm{ }}\left( {{x^2} - xy + {y^2}} \right) - {x^2}\left( {x - y} \right)\\
= {x^3} + {y^3} - {x^3} + {x^2}y\\
= {y^3} + {x^2}y\\
{\rm{ }}C){\rm{ }}\left( {2{x^3} - 9{x^2} + 11x - 3} \right):\left( {2x - 3} \right) - \left( {{x^3} - 3{x^2}} \right):x\\
= \left( {2{x^3} - 3{x^2} - 6{x^2} + 9x + 2x - 3} \right):\left( {2x - 3} \right) - x\left( {{x^2} - 3x} \right):x\\
= \left[ {{x^2}\left( {2x - 3} \right) - 3x\left( {2x - 3} \right) + \left( {2x - 3} \right)} \right]:\left( {2x - 3} \right) - \left( {{x^2} - 3x} \right)\\
= \left( {2x - 3} \right)\left( {{x^2} - 3x + 1} \right):\left( {2x - 3} \right) - {x^2} + 3x\\
= {x^2} - 3x + 1 - {x^2} + 3x\\
= 1\\
{\rm{ }}D){\rm{ }}\left( {{x^3} - 8{y^3} + 4{y^2} - {x^2} + x - 2y} \right):\left( {x - 2y} \right)\\
= \left[ {\left( {x - 2y} \right)\left( {{x^2} + 2xy + 4{y^2}} \right) - \left( {x - 2y} \right)\left( {x + 2y} \right) + \left( {x - 2y} \right)} \right]:\left( {x - 2y} \right)\\
= \left( {x - 2y} \right)\left( {{x^2} + 2xy + 4{y^2} - x - 2y + 1} \right):\left( {x - 2y} \right)\\
= {x^2} + 2xy + 4{y^2} - x - 2y + 1
\end{array}\]

...Xem thêm

Answer 2

\[\begin{array}{l}
\;A){\rm{ }}{\left( {2x - 3} \right)^2} + \left( {3x + 1} \right)\left( {4 - 2x} \right)\\
= 4{x^2} - 12x + 9 + 12x - 6{x^2} + 4 - 2x\\
= - 2{x^2} - 2x + 13\\
{\rm{ }}B){\rm{ }}\left( {x + y} \right){\rm{ }}\left( {{x^2} - xy + {y^2}} \right) - {x^2}\left( {x - y} \right)\\
= {x^3} + {y^3} - {x^3} + {x^2}y\\
= {y^3} + {x^2}y\\
{\rm{ }}C){\rm{ }}\left( {2{x^3} - 9{x^2} + 11x - 3} \right):\left( {2x - 3} \right) - \left( {{x^3} - 3{x^2}} \right):x\\
= \left( {2{x^3} - 3{x^2} - 6{x^2} + 9x + 2x - 3} \right):\left( {2x - 3} \right) - x\left( {{x^2} - 3x} \right):x\\
= \left[ {{x^2}\left( {2x - 3} \right) - 3x\left( {2x - 3} \right) + \left( {2x - 3} \right)} \right]:\left( {2x - 3} \right) - \left( {{x^2} - 3x} \right)\\
= \left( {2x - 3} \right)\left( {{x^2} - 3x + 1} \right):\left( {2x - 3} \right) - {x^2} + 3x\\
= {x^2} - 3x + 1 - {x^2} + 3x\\
= 1\\
{\rm{ }}D){\rm{ }}\left( {{x^3} - 8{y^3} + 4{y^2} - {x^2} + x - 2y} \right):\left( {x - 2y} \right)\\
= \left[ {\left( {x - 2y} \right)\left( {{x^2} + 2xy + 4{y^2}} \right) - \left( {x - 2y} \right)\left( {x + 2y} \right) + \left( {x - 2y} \right)} \right]:\left( {x - 2y} \right)\\
= \left( {x - 2y} \right)\left( {{x^2} + 2xy + 4{y^2} - x - 2y + 1} \right):\left( {x - 2y} \right)\\
= {x^2} + 2xy + 4{y^2} - x - 2y + 1
\end{array}\]

1 câu trả lời 299

Câu hỏi hot cùng chủ đề

Bài viết mới nhất Lớp 8